Evaluate Division

题目描述：

ou are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [Ai, Bi] and values[i] represent the equation Ai / Bi = values[i]. Each Ai or Bi is a string that represents a single variable.

You are also given some queries, where queries[j] = [Cj, Dj] represents the jth query where you must find the answer for Cj / Dj = ?.

Return the answers to all queries. If a single answer cannot be determined, return -1.0.

Note: The input is always valid. You may assume that evaluating the queries will not result in division by zero and that there is no contradiction.

Example 1:

Input: equations = [[“a”,”b”],[“b”,”c”]], values = [2.0,3.0], queries = [[“a”,”c”],[“b”,”a”],[“a”,”e”],[“a”,”a”],[“x”,”x”]]
Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000]
Explanation:
Given: a / b = 2.0, b / c = 3.0
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ?
return: [6.0, 0.5, -1.0, 1.0, -1.0 ]
Example 2:

Input: equations = [[“a”,”b”],[“b”,”c”],[“bc”,”cd”]], values = [1.5,2.5,5.0], queries = [[“a”,”c”],[“c”,”b”],[“bc”,”cd”],[“cd”,”bc”]]
Output: [3.75000,0.40000,5.00000,0.20000]
Example 3:

Input: equations = [[“a”,”b”]], values = [0.5], queries = [[“a”,”b”],[“b”,”a”],[“a”,”c”],[“x”,”y”]]
Output: [0.50000,2.00000,-1.00000,-1.00000]

Constraints:

1 <= equations.length <= 20
equations[i].length == 2
1 <= Ai.length, Bi.length <= 5
values.length == equations.length
0.0 < values[i] <= 20.0
1 <= queries.length <= 20
queries[i].length == 2
1 <= Cj.length, Dj.length <= 5
Ai, Bi, Cj, Dj consist of lower case English letters and digits.

思路:

可以两种方法做，并查集和bfs，把a/b =2 看成 a->b = 2,b->a = 1/2,用图的思想做。

代码:

func calcEquation(equations [][]string, values []float64, queries [][]string) []float64 {
    
    var res []float64
    graph := map[string]map[string]float64{}
    for k,v := range equations{
        if _,ok := graph[v[0]];!ok{
            graph[v[0]] = make(map[string]float64, 0)
        }
        if _,ok := graph[v[1]];!ok{
            graph[v[1]] = make(map[string]float64, 0)
        }
        graph[v[0]][v[1]] = values[k]
        graph[v[1]][v[0]] = 1/values[k]
        graph[v[0]][v[0]] = 1
        graph[v[1]][v[1]] = 1
    }
    var calc func(a, b string, visit map[string]bool)float64
    calc = func(a,b string, visit map[string]bool)float64{
        if _,ok := graph[a];!ok{
            return -1
        }
        if _,ok := graph[b];!ok{
            return -1
        }
        if v,ok := graph[a][b];ok{
            return v
        }else{
            for key,value := range graph[a]{
                // 防止重复
                if _,ok := (visit)[key];ok{
                    continue
                }
                visit[a] = true
                ans := calc(key,b,visit)
                if ans != -1{
                    return ans *value
                }
            }
        }
        return -1
        
    }
    for _,v := range queries{
        res = append(res, calc(v[0],v[1], map[string]bool{}))
    }
    return res
}

代码效率：

执行用时：0 ms, 在所有 Go 提交中击败了100.00%的用户
内存消耗：2.2 MB, 在所有 Go 提交中击败了100.00%的用户